Can we solve this equation
$$\left(1+\frac{x}{b}\right)e^{-x/b}=z$$
We have to determine value of $x$ in term of $z$.
Problem occur while calculating the following integral.
$$\frac{a}{b^{2}}\int_{0}^{\infty}(1-(1+\frac{x}{b})e^{-\frac{x}{b}})^{a-1}x^{r+1}e^{-\frac{x}{b}}dx $$ and subtituting $$(1+\frac{x}{b})e^{-\frac{x}{b}}=z$$
On
Your original problem is $f(a, b, r) =\frac{a}{b^{2}}\int_{0}^{\infty}(1-(1+\frac{x}{b})e^{-\frac{x}{b}})^{a-1}x^{r+1}e^{-\frac{x}{b}}dx $.
I'll play with it and see what I get.
Letting $x/b = y$, so $dx = b dy$, $f(a, b, r) =\frac{a}{b}\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}(by)^{r+1}e^{-y}dy =ab^r\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}y^{r+1}e^{-y}dy $.
Since $e^y > 1+y$ for $y ? 0$, $(1+y)e^{-y} < 1$ so we could expand $(1-(1+y)e^{-y})^{a-1} $ with the binomial theorem. This gives $(1-(1+y)e^{-y})^{a-1} =\sum_{n=0}^{\infty} (-1)^n \binom{a-1}{n} ((1+y)e^{-y})^n $.
Integrating term-by-term,
$\begin{array}\\ f(a, b, r) &=ab^r\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}y^{r+1}e^{-y}dy\\ &=ab^r\int_{0}^{\infty}\sum_{n=0}^{\infty} (-1)^n \binom{a-1}{n} ((1+y)e^{-y})^ny^{r+1}e^{-y}dy\\ &=ab^r\sum_{n=0}^{\infty}(-1)^n \binom{a-1}{n} \int_{0}^{\infty} ((1+y)e^{-y})^ny^{r+1}e^{-y}dy\\ \end{array} $
Looking at the inside integral,
$\begin{array}\\ \int_{0}^{\infty} ((1+y)e^{-y})^ny^{r+1}e^{-y}dy &=\int_{0}^{\infty} (1+y)^ne^{-ny}y^{r+1}e^{-y}dy\\ &=\int_{0}^{\infty} \sum_{k=0}^n \binom{n}{k}y^ke^{-(n+1)y}y^{r+1}dy\\ &=\sum_{k=0}^n \binom{n}{k}\int_{0}^{\infty} e^{-(n+1)y}y^{k+r+1}dy\\ &=\sum_{k=0}^n \binom{n}{k}\int_{0}^{\infty} e^{-z}\left(\frac{z}{n+1}\right)^{k+r+1}\frac{dz}{n+1}\quad (z = (n+1)y)\\ &=\sum_{k=0}^n \binom{n}{k}\frac{1}{(n+1)^{k+r+2}}\int_{0}^{\infty} e^{-z}z^{k+r+1}dz\\ &=\sum_{k=0}^n \binom{n}{k}\frac{(k+r+1)!}{(n+1)^{k+r+2}}\\ &=\frac1{(n+1)^{r+2}}\sum_{k=0}^n \binom{n}{k}\frac{(k+r+1)!}{(n+1)^{k}}\\ \end{array} $
There might be something that can be done if this is inserted into the sum above, but I'll leave it at this.
As Famous Blue Raincoat commented, you can rewrite $$\left(1+\frac{x}{b}\right)e^{-x/b}=e\left(1+\frac{x}{b}\right)e^{-(1+x/b)}$$ and setting $y=-(1+\frac{x}{b})$, the equation write $$y e^{y}=-\frac{z}{e}$$ the solution of which being given by Lambert function $$y=W\left(-\frac{z}{e}\right)$$ and, back to $x$, $$x=-b \left(1+W\left(-\frac{z}{e}\right)\right)$$
In a more general manner, any equation which can be written as $$A+B x+C \log(D+Ex)=0$$ has solutions which can expressed in terms of Lambert function.
However, may I confess that I do not see how to compute the integral ?